Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;

public class A815{

	void solve()
	{
		int n = ni(), m = ni();
		int[][] a1 = new int[n][m];
		int[][] a2 = new int[m][n];
		
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
				a1[i][j] = a2[j][i] = ni();
		
		StringBuilder sb1 = new StringBuilder();
		StringBuilder sb2 = new StringBuilder();
		
		int x1 = getAns(a1, n, m, "row ", "col ", sb1);
		int x2 = getAns(a2, m, n, "col ", "row ", sb2);
		
		if(x1 == -1 && x2 == -1)
			out.println(-1);
		else
		{
			if(x1 < x2)
			{
				out.println(x1);
				out.println(sb1);
			}
			else
			{
				out.println(x2);
				out.println(sb2);
			}
		}
		
	}
	
	int getAns(int[][] a, int n, int m, String s1, String s2, StringBuilder sb)
	{
		int[][] c = new int[n][m];
		int ans = 0;
		for(int i=0;i<n;i++)
		{
			int min = -1>>>1;
			for(int x : a[i])
				min = Math.min(min, x);
			for(int j=0;j<m;j++)
				c[i][j] = min;
			if(min != 0)
				for(int p=0;p<min;p++,ans++)
					sb.append(s1 + (i+1) + "\n");
		}
		for(int j=0;j<m;j++)
		{
			int dif = a[0][j] - c[0][j];
			if(dif < 0)
				return -1;
			for(int i=1;i<n;i++)
				if(a[i][j] - c[i][j] != dif)
					return -1;
			for(int p=0;p<dif;p++,ans++)
				sb.append(s2 + (j+1) + "\n");
		}
		return ans;
	}
	
	public static void main(String[] args){new A815().run();}
	
	private byte[] bufferArray = new byte[1024];
	private int bufLength = 0;
	private int bufCurrent = 0;
	InputStream inputStream;
	PrintWriter out;
	
	public void run()
	{
		inputStream = System.in;
		out = new PrintWriter(System.out);
		solve();
		out.flush();
	}
	
	int nextByte()
	{
		if(bufLength==-1)
			throw new InputMismatchException();
		if(bufCurrent>=bufLength)
		{
			bufCurrent = 0;
			try
			{bufLength = inputStream.read(bufferArray);}
			catch(IOException e)
			{ throw new InputMismatchException();}
			if(bufLength<=0)
				return -1;
		}
		return bufferArray[bufCurrent++];
	}
	
	boolean isSpaceChar(int x)
	{return (x<33 || x>126);}
	
	boolean isDigit(int x)
	{return (x>='0' && x<='9');}
	
	int nextNonSpace()
	{
		int x;
		while((x=nextByte())!=-1 && isSpaceChar(x));
		return x;
	}
	
	int ni()
	{
		long ans = nl();
		if ( Integer.MIN_VALUE <= ans && ans <= Integer.MAX_VALUE )
			return (int)ans;
		throw new InputMismatchException();
	}
	
	long nl()
	{
		long ans = 0;
		boolean neg = false;
		int x = nextNonSpace();
		if(x=='-') 
		{
			neg = true;
			x = nextByte();
		}
		while(!isSpaceChar(x))
		{
			if(isDigit(x))
			{
				ans = ans*10 + x -'0';
				x = nextByte();
			}
			else
				throw new InputMismatchException();
		}
		return neg ? -ans:ans;
	}
	
	String ns()
	{
		StringBuilder sb = new StringBuilder();
		int x = nextNonSpace();
		while(!isSpaceChar(x))
		{
			sb.append((char)x);
			x = nextByte();
		}
		return sb.toString();
	}
	
	char nc()
	{ return (char)nextNonSpace();}
	
	double nd()
	{ return (double)Double.parseDouble(ns()); }
	
	char[] ca()
	{ return ns().toCharArray();}
	
	char[] ca(int n)
	{
		char[] ans = new char[n];
		int p =0;
		int x = nextNonSpace();
		while(p<n)
		{
			ans[p++] = (char)x;
			x = nextByte();
		}
		return ans;
	}
	
	int[] ia(int n)
	{
		int[] ans = new int[n];
		for(int i=0;i<n;i++)
			ans[i]=ni();
		return ans;
	}
	
}